Difference between revisions of "2022 AMC 8 Problems/Problem 12"

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<math>\textbf{(A)} ~\dfrac{1}{16}\qquad\textbf{(B)} ~\dfrac{1}{8}\qquad\textbf{(C)} ~\dfrac{1}{4}\qquad\textbf{(D)} ~\dfrac{3}{8}\qquad\textbf{(E)} ~\dfrac{1}{2}</math>
 
<math>\textbf{(A)} ~\dfrac{1}{16}\qquad\textbf{(B)} ~\dfrac{1}{8}\qquad\textbf{(C)} ~\dfrac{1}{4}\qquad\textbf{(D)} ~\dfrac{3}{8}\qquad\textbf{(E)} ~\dfrac{1}{2}</math>
  
==Solution==
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==Solution 1==
  
 
First, we realize that there are a total of <math>16</math> possibilities. Now, we list all of them that can be spun. This includes <math>64</math> and <math>81</math>. Then, our answer is <math>\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}</math>.
 
First, we realize that there are a total of <math>16</math> possibilities. Now, we list all of them that can be spun. This includes <math>64</math> and <math>81</math>. Then, our answer is <math>\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}</math>.
  
 
~MathFun1000
 
~MathFun1000
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 +
==Solution 2==
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There are <math>4 \cdot 4 = 16</math> total possibilities of <math>N</math>. We know <math>N=10A+B</math>, which <math>A</math> is a number from spinner <math>A</math>, and <math>B</math> is a number from spinner <math>B</math>. Also, notice that there are no perfect squares in the <math>50</math>s or <math>70</math>s, so only  <math>4-2=2</math> values of N work, namely <math>64</math> and <math>81</math>. Hence, <math>\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}</math>.
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~MrThinker
  
 
==Video Solution==
 
==Video Solution==

Revision as of 09:24, 7 October 2022

Problem

The arrows on the two spinners shown below are spun. Let the number $N$ equal $10$ times the number on Spinner $\text{A}$, added to the number on Spinner $\text{B}$. What is the probability that $N$ is a perfect square number? [asy] //diagram by pog give me 1 billion dollars for this size(6cm); usepackage("mathptmx"); filldraw(arc((0,0), r=4, angle1=0, angle2=90)--(0,0)--cycle,mediumgray*0.5+gray*0.5); filldraw(arc((0,0), r=4, angle1=90, angle2=180)--(0,0)--cycle,lightgray); filldraw(arc((0,0), r=4, angle1=180, angle2=270)--(0,0)--cycle,mediumgray); filldraw(arc((0,0), r=4, angle1=270, angle2=360)--(0,0)--cycle,lightgray*0.5+mediumgray*0.5); label("$5$", (-1.5,1.7)); label("$6$", (1.5,1.7)); label("$7$", (1.5,-1.7)); label("$8$", (-1.5,-1.7)); label("Spinner A", (0, -5.5)); filldraw(arc((12,0), r=4, angle1=0, angle2=90)--(12,0)--cycle,mediumgray*0.5+gray*0.5); filldraw(arc((12,0), r=4, angle1=90, angle2=180)--(12,0)--cycle,lightgray); filldraw(arc((12,0), r=4, angle1=180, angle2=270)--(12,0)--cycle,mediumgray); filldraw(arc((12,0), r=4, angle1=270, angle2=360)--(12,0)--cycle,lightgray*0.5+mediumgray*0.5); label("$1$", (10.5,1.7)); label("$2$", (13.5,1.7)); label("$3$", (13.5,-1.7)); label("$4$", (10.5,-1.7)); label("Spinner B", (12, -5.5)); [/asy] $\textbf{(A)} ~\dfrac{1}{16}\qquad\textbf{(B)} ~\dfrac{1}{8}\qquad\textbf{(C)} ~\dfrac{1}{4}\qquad\textbf{(D)} ~\dfrac{3}{8}\qquad\textbf{(E)} ~\dfrac{1}{2}$

Solution 1

First, we realize that there are a total of $16$ possibilities. Now, we list all of them that can be spun. This includes $64$ and $81$. Then, our answer is $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$.

~MathFun1000

Solution 2

There are $4 \cdot 4 = 16$ total possibilities of $N$. We know $N=10A+B$, which $A$ is a number from spinner $A$, and $B$ is a number from spinner $B$. Also, notice that there are no perfect squares in the $50$s or $70$s, so only $4-2=2$ values of N work, namely $64$ and $81$. Hence, $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$.

~MrThinker

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1008

~Interstigation

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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